Question: The radius of a sphere is increasing at a rate of $7.5$ meters per minute. At a certain instant, the radius is $5$ meters. What is the rate of change of the surface area of the sphere at that instant (in square meters per minute)? Choose 1 answer: Choose 1 answer: (Choice A) A $300\pi$ (Choice B) B $100\pi$ (Choice C) C $225\pi$ (Choice D) D $330\pi$ The surface area of a sphere with radius $r$ is $4\pi r^2$.
Answer: Setting up the math Let... $r(t)$ denote the sphere's radius at time $t$, and $S(t)$ denote the sphere's surface area at time $t$. We are given that $r'(t)=7.5$ and that $r(t_0)=5$ for a specific time $t_0$. We want to find $S'(t_0)$. Relating the measures $S(t)$ and $r(t)$ relate to each other through the formula for the surface area of a sphere: $S(t)=4\pi[r(t)]^2$ We can differentiate both sides to find an expression for $S'(t)$ : $S'(t)=8\pi r(t)r'(t)$ Using the information to solve Let's plug ${r(t_0)}={5}$ and ${r'(t_0)}={7.5}$ into the expression for $S'(t_0)$ : $\begin{aligned} S'(t_0)&=8\pi{r(t_0)}{r'(t_0)} \\\\ &=8\pi({5})({7.5}) \\\\ &=300\pi \end{aligned}$ In conclusion, the rate of change of the surface area of the sphere at that instant is $300\pi$ square meters per minute. Since the rate of change is positive, we know that the surface area is increasing.